Part Three: Thermodynamics

Steam turbine converts the energy of high-pressure steam to mechanical energy and then electricity.

World energy consumption (2008): ~ 151012 (15 trillion / tera) watts., mostly from fossil fuels

Applications of the laws of thermodynamics:

• Combustion engines.

• Sun Heat flows on Earth Climate.

• Global warming.

• Big bang: Heat flow in the universe.

Rank   Country  Electricity consumption (MW·h/yr)  

Year of Data   Source   Population   As of  Average power per capita (watts per person)  

—  World 16,830,000,000 2005 CIA Est.[3] 6,464,750,000 2005 297

1  United States 3,816,000,000 2005 CIA[1] 298,213,000 2005 1,460

2  China 3,640,000,000 2009 [4] 1,315,844,000 2009 277

—  European Union[5] 2,820,000,000 2004 CIA Est. 459,387,000 2005 700

3  Russia 985,200,000 2007 CIA Est. 143,202,000 2005 785

4  Japan 974,200,000 2005 CIA 128,085,000 2005 868

5  Germany 593,400,000 2007 CIA Est. 82,329,758 2009 (CIA Est.) 822.22

6  Canada 540,200,000 2005 CIA 32,268,000 2005 1,910

7  India 488,500,000 2005 CIA 1,103,371,000 2005 50.5

8  France 451,500,000 2005 CIA 60,496,000 2005 851

9  South Korea 368,600,000 2007 CIA 47,817,000 2005 879

10  Brazil 368,500,000 2005 CIA 186,405,000 2005 226

11  United Kingdom 348,700,000 2005 CIA 59,668,000 2005 667

12  Italy 307,100,000 2005 CIA 58,093,000 2005 603

13  Spain 243,000,000 2005 CIA 43,064,000 2005 644

14  South Africa 241,400,000 2007 CIA 47,432,000 2005 581

15  Taiwan (Republic of China) 221,000,000 2006 CIA 22,894,384 2005 1,101

16  Australia 219,800,000 2005 CIA 20,155,000 2005 1,244

17  Mexico 183,300,000 2005 CIA 107,029,000 2005 195

List of countries by electricity consumption

Part Three: Thermodynamics

16. Temperature and Heat

17. The Thermal Behavior of Matter

18. Heat, Work, and the First Law of Thermodynamics

19. The Second Law of Thermodynamics

16. Temperature & Heat

1. Heat , Temperature & Thermodynamic Equilibrium

2. Heat Capacity & Specific Heat

3. Heat Transfer

4. Thermal Energy Balance

How does this photo reveal heat loss from the

house?

And how can you tell that the car was recently

driven?

Studies of thermal properties:

• Thermodynamics: Relations between macroscopic properties.

• Statistical mechanics: Atomic description.

IR photo: engine & brakes hot

16.1. Heat , Temperature & Thermodynamic Equilibrium

Thermodynamic equilibrium:State at which macroscopic properties of system remains unchanged over time.

Examples of macroscopic properties:

L, V, P, , , …

0th law of thermodynamics:2 systems in thermodynamic equilibrium with a 3rd system are themselves in equilibrium.

2 systems are in thermal contact if heating one of them changes the other.

Otherwise, they are thermally insulated.

Two systems have the same temperature

they are in thermodynamic equilibrium

A,B in eqm

B,C in eqm A,C in eqm

Gas Thermometers & the Kelvin Scale

Constant volume gas thermometer T P

Kelvin scale:

P = 0 0 K = absolute zero

Triple point of water 273.16 K

Triple point: T at which solid, liquid & gas phases co-exist in equilibrium

All gases behave similarly as P 0.

Mercury fixed at this level by adjusting h P T.

Temperature Scales

Celsius scale ( C ) :

Melting point of ice at P = 1 atm TC = 0 C.

Boiling point of water at P = 1 atm TC = 100 C.

Triple point of water = 0.01C

273.15CT T CT T

Fahrenheit scale ( F ) :

Melting point of ice at P = 1 atm TF = 32 F.

Boiling point of water at P = 1 atm TF = 212 F.180

32100F CT T 9

5F CT T

Rankine scale ( R ) :

0 0R K

R FT T

Supplement

P, V, n, T

Conditions for thermodynamic equilibrium

P V = n R TIsolated ideal gas

P1 , V1 , n1 , T1

P2 , V2 , n2 , T2

Fixed, thermally conducting partition

P j V j = n j R T j

T 1 = T 2

j = 1, 2

Movable, thermally conducting partition

P j V j = n j R T j

T 1 = T 2 & P 1 = P 2

j = 1, 2

Porous, movable, thermally conducting partition

P j V j = n j R T j

T 1 = T 2 , P 1 = P 2 & 1 = 2

j = 1, 2

= n / VSame as no partition

(Local eqm.)

P1 , V1 , n1 , T1

P2 , V2 , n2 , T2

P1 , V1 , n1 , T1

P2 , V2 , n2 , T2

Heat & Temperature

Brief history of the theory of heat:

1.Heat is a fluid (caloric theory: 1770s) that flows from hot to cold bodies.

2.B.Thompson, or Count Rumford, (late 1790s): unlimited amount of heat

can be produced in the boring of canon heat is not conserved.

3.J.Joule (1840s): Heat is a form of energy.

A match will burn your finger, but doesn’t provide much heat.

Heat ~ amount

Temperature ~ intensity

Heat is energy transferred from high to low temperature regions.

16.2. Heat Capacity & Specific Heat

Q C T Heat capacity C of a body:

Q = heat transferred to body. /C J K

Specific heat c = heat capacity per unit mass Q m c T

/c J kg K

1 calorie (cal) = heat needed to raise 1 g of water from 14.5C to 15.5C.

1 BTU = heat needed to raise 1 lb of water from 63F to 64F.

1 4.184

1 1054

cal J

BTU J

c = c(P,V) for gases cP , cV .

Example 16.1. Waiting to Shower

The temperature in the water heater has dropped to 18C.

If the heater holds 150 kg of water, how much energy will it take to bring it up to 50C?

If the energy is supplied by a 5.0 kW electric heating element, how long will that take?

Q m c T 150 4184 / 50 18kg J kg K C C 20 MJ

Qt

P

20

5.0

MJ

kW

6

3

20 10

5.0 10 /

J

J s

4000 s 1 7 mh

The Equilibrium Temperature

Heat flows from hot to cold objects until a common equilibrium temperature is reached.

For 2 objects insulated from their surroundings:

1 2 0Q Q 1 1 1 2 2 2m c T m c T

When the equilibrium temperature T is reached:

1 1 1 2 2 2 0m c T T m c T T

1 1 1 2 2 2

1 1 2 2

m c T m c TT

m c m c

GOT IT? 16.1.

A hot rock with mass 250 g is dropped into an equal mass of pool water.

Which temperature changes more? Explain.

crock = 0.20 cal / g C

cwater = 1.0 cal / g C Trock changes more

Example 16.2. Cooling Down

An aluminum frying pan of mass 1.5 kg is at 180C,

when it was plunged into a sink containing 8.0 kg of water at 20C.

Assuming none of the water boils & no heat is lost to the environment,

find the equilibrium temperature of the water & pan.

1 1 1 2 2 2

1 1 2 2

m c T m c TT

m c m c

1.5 900 / 180 8.0 4184 / 20

1.5 900 / 8.0 4184 /

kg J kg K C kg J kg K C

kg J kg K kg J kg K

26 C

16.3. Heat Transfer

Common heat-transfer mechanisms:

• Conduction

• Convection

• Radiation

Conduction

Conduction: heat transfer through direct physical contact.

Mechanism: molecular collision.

Thermal conductivity k ,

[ k ] = W / mK

dQH

dt

Tk A

x

Heat flow H , [ H ] = watt :

conductor

insulator

Example 16.3. Warming a Lake

A lake with flat bottom & steep sides has surface area 1.5 km2 & is 8.0 m deep.

The surface water is at 30C; the bottom, 4.0C.

What is the rate of heat conduction through the lake?

Assume T decreases uniformly from surface to bottom.

6 2 30 4.00.61 / 1.5 10

8.0

C CW m m

mK

TH k A

x

3.0 MW

Power of sunlight is ~ 1 kW / m2 .

TH k A

x

applies only when T = const over each (planar) surface

For complicated surface, use d T

H k Ad x

Prob. 72 & 78.

Composite slab:

H must be the same in both slabs to

prevent accumulated heat at interface

3 22 11 2

1 2

T TT TH k A k A

x x

Thermal resistance :x

Rk A

[ R ] = K / W

2 1

1

T T

R

1 3

1 2

T TH

R R

1 2 1T T H R

2 3 2T T H R Resistance in series

TH

R

3 2

2

T T

R

GOT IT? 16.2.

Rank order the 3 temperature differences.

1 3 2T T T

H, A, x same for all three k T = const

Insulating properties of building materials are described by the -factor ( -value ) .

xR A

k

R = thermal resistance of a slab of unit area

2 /m K WR

2 /ft F h BTU RU.S.

TA

H

2 21 / 0.176 /ft F h BTU m K W

d TH k A

d x

T

R

AT

R

Example 16.4. Cost of Oil

The walls of a house consist of plaster ( = 0.17 ), -11 fiberglass

insulation, plywood ( = 0.65 ), and cedar shingles ( = 0.55 ).

The roof is the same except it uses -30 fiberglass insulation.

In winter, average T outdoor is 20 F, while the house is at 70 F.

The house’s furnace produces 100,000 BTU for every gallon of oil,

which costs $2.20 per gallon. How much is the monthly cost?

0.17 11 0.65 0.55wall R 12.370.17 30 0.65 0.55roof R 31.37

2 36 28 10rectA ft ft ft 21280 ft21164 ft 14

2 36cos30roof

ftA ft 1

2 28 14 tan 302gableA ft ft 2226 ft

21506wall rect gableA A A ft

2 21/ / / 1506 70 20

12.37wallH BTU h ft F ft F F

2 21/ / / 1164 70 20

31.37roofH BTU h ft F ft F F

6073 /BTU h

1853 /BTU h

6073 1853 / 24 / 30 /Q BTU h h d d month 5.7 MBTU

5.7 10 / $ 2.20 /Cost MBTU gal MBTU gal $126

Convection

Convection = heat transfer by fluid motion

T rises

Convection cells in liquid film between glass plates(Rayleigh-Bénard convection, Benard cells)

Examples:

• Boiling water.

• Heating a house.

• Sun heating earth Climate, storms.

• Earth mantle continental drift

• Generation of B in stars & planets.

convecH T

Radiation

Glow of a stove burner it loses energy by radiation

4Pe T

AStefan-Boltzmann law for radiated power:

= Stefan-Boltzmann constant = 5.67108 W / m2 K4.

A = area of emitting surface.

0 < e < 1 is the emissivity ( effectiveness in emitting radiation ).

e = 1 perfect emitter & absorber ( black body ).

Black objects are good emitters & absorbers.

Shiny objects are poor emitters & absorbers.

Wien‘s displacement law : max = b / T

P T4 Radiation dominates at high T.

Wavelength of peak radiation becomes shorter as T increases.

Sun ~ visible light.

Near room T ~ infrared.

4Pe T

AStefan-Boltzmann law :

sun sunRT

RT

T

T

32.898 10b mK

.502 5778

300

m K

K

9.66 m

GOT IT? 16.3.

Name the dominant form of heat transfer from

(a) a red-hot stove burner with nothing on it.

(b) a burner in direct contact with a pan of water.

(c) the bottom to the top of the water in the pan once it boils.

Radiation

conduction

convection

Example 16.5. Sun’s Temperature

The sun radiates energy at the rate P = 3.91026 W, & its radius is 7.0 108

m.

Treating it as a blackbody ( e = 1 ), find its surface temperature.4P e AT

= 5.67108 W / m2 K4

1/4

26

28 2 4 8

3.9 10

5.67 10 / 4 7.0 10

W

W m K m

1/4

24

PT

e R

35.8 10 K

16.4. Thermal Energy Balance

A house in thermal-energy balance.

System with fixed rate of energy input

tends toward an energy- balanced state

due to negative feedback.

Heat from furnace balances

losses thru roofs & walls

Example 16.6. Hot Water

A poorly insulated electric water heater loses heat by conduction at the rate of

120 W for each C difference between the water & its surrounding.

It’s heated by a 2.5 kW heating element & is located in a basement kept at 15 C.

What’s the water temperature if the heating element operates continuously.

120 / 15H W C T C 2.5 kW

32.5 1015

120 /

WT C

W C

36 C

Conductive heat loss

Electrical energy in

Heating element

T = ?

Example 16.7. Solar Greenhouse

A solar greenhouse has 300 ft2 of opaque -30 walls,

& 250 ft2 of -1.8 double-pane glass that admits solar energy at the rate of 40 BTU / h / ft2.

Find the greenhouse temperature on a day when outdoor temperature is 15 F.

T A TH

R

R

67 F

2

2

300

30 /wall

ft TH

ft F h BTU

10 / /BTU h F T

2

2

250

1.8 /glass

ft TH

ft F h BTU

139 / /BTU h F T

2 240 / / 250sunH BTU h ft ft 410 /BTU h

410 /

149 / /

BTU hT

BTU h F

15 67T F F 82 F

wall glassH H

Application: Greenhouse Effect & Global Warming

Average power from sun :

2960 /S W m

Total power from sun : 2S EH S R

Power radiated (peak at IR) from Earth :

2 44E EH e R T

S EH H

18 C

1/42

8 2 4

960 /

5.67 10 / 4

W mT

W m K

255 K1e

C.f. T 15 C natural greenhouse effect

Greenhouse gases: H2O, CO2 , CH4 , …

passes incoming sunlight, absorbs outgoing IR.

Mars: none

Venus: huge

CO2 increased by 36%

0.6 C increase during 20th century.

1.5 C – 6 C increase by 2100.